Question: $\text C = \left[\begin{array}{r}-1 \\ -2 \\ 2\end{array}\right]$ and $\text D = \left[\begin{array}{rr}2 & 1\end{array}\right]$ Let $\text {H = CD}$. Find $\text H$. $ {H = }$
Answer: The Strategy When multiplying matrices, we should find each entry of the resulting product matrix separately. To find entry $(i,j)$ of the resulting product matrix, we calculate the vector dot product of row $i$ of the first matrix and column $j$ of the second matrix. [I don't know what "vector dot product" is!] Finding $\text {H}_{1,1}$ $\text{H}_{1,1}$ is the dot product of the first row of $\text{C}$ and the first column of $\text{D}$. $ \text {H}=\left[\begin{array}{rr}{-1} \\ -2 \\ 2\end{array}\right]\left[\begin{array}{rr} {2} & 1\end{array}\right]$ Therefore, this is the appropriate calculation of $\text{H}_{1,1}$. $\begin{aligned}\text{H}_{1,1}&=(-1)\cdot(2)\\\\ &=-2 \end{aligned}$ The other entries of $\text{H}$ can be found similarly. Try it yourself for $\text{H}_{2,1}$ What is the appropriate calculation of ${H}_{2,1}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-2 \cdot 1 = -2$ (Choice B) B $-2 \cdot 2 = -4$ (Choice C) C ${H}_{2,1}$ does not exist. Check Summary After calculating all the remaining entries of $\text{H}$, we get the following answer. $ \text {H}=\left[\begin{array}{rr}-2 & -1 \\ -4 & -2 \\ 4 & 2\end{array}\right]$